ok, i'm stuck with this problem and I have to use a website to turn in the answer that is limited to say if it is right or wrong, which doesn't help me at all.
A uniform board of length L and mass M lies near a boundary that separates two regions. In region 1, the coefficient of kinetic friction between the board and the surface is ų1, and in region 2, the coefficient is ų2. The positive direction is shown in the figure.
http://session.masteringphysics.com/problemAsset/1011078/29/MWE_cw_5_a.jpg
Find the net work W done by friction in pulling the board directly from region 1 to region 2. Assume that the board moves at constant velocity.
Express the net work in terms of M, g, L, ų1, and ų2.
Any ideas? Thank you in advance.
2007-03-05 18:02:26 · 3 answers · asked by Sergio__ 7 in Science & Mathematics ➔ Physics
W= Force * Distance
You must overcome only the force of friction so the work required to move the object will be equal to the work done by friction
Force of friction= Normal force times the coefficient = (M*g)*u
Distance = L
so
Work done by friction= - [(M)(g)(u1)(L)]+[(M)(g)(u2)(L)]
the work done by friction is negative because the friction is acting in the opposite direction of the movement.
2007-03-05 18:46:32 · answer #1 · answered by ThE_HooLiGaN 3 · 0⤊ 3⤋
W=-(MgL/2)(u1+u2)
2015-11-27 14:55:21 · answer #2 · answered by Yevgeniy Chubarov 1 · 2⤊ 0⤋
ThE_HooLiGaN almost had it...
W=integral(0 to L) F(x)dx
W=(b^2-a^2)/2
...
W=(1/2)[(Mgu2)L-(Mgu1)L]
2015-10-11 16:13:38 · answer #3 · answered by ? 2 · 2⤊ 0⤋